Question

我需要根据n个输入列表获得所有可能的组合，并为它们做一些事情。

当前代码示例：

import itertools

# example inputs
list_small = [1, 2, 3]
list_medium = [444, 666, 242]
list_huge = [1680, 7559, 5573, 43658, 530, 11772, 284, 50078, 783, 37809, 6740, 37765, 74492, 50078, 783, 37809, 6740, 37765, 74492]

# out of the input list, I need to generate all numbers from 0 to the current list element
# e.g. if I have 6, I need to get [0, 1, 2, 3, 4, 5, 6]
# if I get a list [1, 2, 3], the output will be [[0, 1], [0, 1, 2], [0, 1, 2, 3]]
# I achieved this by doing it with xrange: [x for x in xrange(0, current_list_element + 1)]
# after that, I need to generate all possible combinations using the generated lists
# I managed to do this by using itertools.product()

# print this to get all possible combinations
# print list(itertools.product(*[[x for x in xrange(0, current_list_element + 1)] for current_list_element in list_medium]))

cumulative_sum = 0
for current_combination in itertools.product(*[[x for x in xrange(0, current_list_element + 1)] for current_list_element in list_medium]):
    # now I need to do some calculations to the current combination
    # e.g. get sum of all combinations, this is just an example
    cumulative_sum += sum(current_combination)

    # another example
    # get XOR sum of current combination, more at https://en.wikipedia.org/wiki/Exclusive_or
    print reduce(operator.xor, current_combination, 0)

# runs fast for list_small, then takes some time for list_medium and then takes ages for list_huge
print cumulative_sum

这适用于较小的列表，但是对于较大的列表/或者抛出运行时错误需要无穷大。有没有更好的方法来做到这一点？获得所有组合的更好方法？或者我是以某种错误的方式使用xrange？

我用Python 2.7和Pypy 2尝试过这个。

编辑：感谢@famagusta我摆脱了xrange，但问题仍然存在

import itertools

# example inputs
list_small = [1, 2, 3]
list_medium = [444, 666, 242]
list_huge = [1680, 7559, 5573, 43658, 530, 11772, 284, 50078, 783, 37809, 6740, 37765, 74492, 50078, 783, 37809, 6740, 37765, 74492]

max_element = max(get_input_stones)
combo_list = range(0, max_element + 1)

cumulative_sum = 0
for current_combination in itertools.product(*combo_list):
    # now I need to do some calculations to the current combination
    # e.g. get sum of all combinations, this is just an example
    cumulative_sum += sum(current_combination)

    # another example
    # get XOR sum of current combination, more at https://en.wikipedia.org/wiki/Exclusive_or
    print reduce(operator.xor, current_combination, 0)

# runs fast for list_small, then takes some time for list_medium and then takes ages for list_huge
print cumulative_sum

Answer 1

生成此类嵌套列表可能会让您遇到内存限制问题。您可以只使用从列表中最大数字生成的一个超级列表，而不是重复生成子列表。只需存储较小元素将停止的索引。

例如，[1,6,10] - [0,1,2,3,4,5,6,7,8,9,10]，[1,6,10]

第二个列表告诉您在第一个列表中停止的位置以提取计算所需的子列表

这应该可以节省一些空间。

list_small = [1, 2, 3]
list_medium = [444, 666, 242]
list_huge = [1680, 7559, 5573, 43658, 530, 11772, 284, 50078, 783, 37809, 6740, 37765, 74492, 50078, 783, 37809, 6740, 37765, 74492]

max_element = max(list_huge)   # being lazy here - write a max function
combo_list = range(0, max_element + 1)  # xrange does not support slicing

cumulative_sum = 0
for element in list_huge:
    cumulative_sum += sum(combo_list[:element])

print(cumulative_sum)

有没有更好的方法来计算多个列表的产品？

1 个答案: